Question

A gas undergoes physical adsorption on a surface and follows the given Freundlich adsorption isotherm equation $\frac{x}{m}=k{p}^{0.5}$
Adsorption of the gas increase with

• A. decrease in $p$ and decrease in $T$
• B. increase in $p$ and increase in $T$
• C. increase in $p$ and decrease in $T$
• D. decrease in $p$ and increase in $T$

Answer 1

The correct option is C increase in $p$ and decrease in $T$

For physisorption or physical adsorption,
Adsorption isotherm (Temperature, $T=\text{constant}$) is shown below:where, $x=\text{amount of adsorbate},m=\text{amount of adsorbent}$,
$\frac{x}{m}=\text{degree of adsorption}$
$\frac{1}{n}=\text{order of the reaction}$, where, $0<\frac{1}{n}<1$ and so, $1<n<\infty$
Here, $\frac{x}{m}=k{p}^{\frac{1}{2}}$,
i.e. $\frac{x}{m}\alpha p^{\frac{1}{2}}$
Adsorption isobar (Pressure, $p=\text{constant}$)
So, the rate of physical adsorption of the gas, increases with $p$ (when, $T$ is constant) and decreases with $T$ (when $p$ is constant).